# 实验任务4

#include <stdio.h>
const int N = 4;
int main()
{
int a[N] = {2, 0, 2, 1};
char b[N] = {'2', '0', '1', '1'};
int i;
printf("sizeof(int) = %d\n", sizeof(int));
printf("sizeof(char) = %d\n", sizeof(char));
printf("\n");

for (i = 0; i < N; ++i)
printf("%x: %d\n", &a[i], a[i]);
printf("\n");

for (i = 0; i < N; ++i)
printf("%x: %c\n", &b[i], b[i]);
return 0;
}

是，4

#include <stdio.h>
int main()
{
int a[2][3] = {{1,2,3},{4,5,6}};
char b[2][3] = {{'1', '2', '3'}, {'4', '5', '6'}};
int i, j;

for (i = 0; i < 2; ++i)
for (j = 0; j < 3; ++j)
printf("%x: %d\n", &a[i][j], a[i][j]);
printf("\n");

for (i = 0; i < 2; ++i)
for (j = 0; j < 3; ++j)
printf("%x: %c\n", &b[i][j], b[i][j]);
return 0;
}

是，4

#include <stdio.h>
#define N 1000
int fun(int n, int m, int bb[N])
{
int i, j, k = 0, flag;
for (j = n; j <= m; j++)
{
flag=1;
for (i = 2; i < j; i++)
if (j%i==0)
{
flag = 0;
break;
}
if (flag!=0)
bb[k++] = j;
}
return k;
}

int main()
{
int n = 0, m = 0, i, k, bb[N];
scanf("%d", &n);
scanf("%d", &m);
for (i = 0; i < m - n; i++)
bb[i] = 0;
k = fun(n,m,bb);
for (i = 0; i < k; i++)
printf("%4d", bb[i]);
return 0;
}

#include <stdio.h>
const int N = 5;
int find_max(int x[], int n);
void input(int x[], int n);
void output(int x[], int n);

int main()
{
int a[N];
int max;
input(a, N);
output(a, N);
max = find_max(a, N);
printf("max = %d\n", max);
return 0;
}

void input(int x[], int n)
{
int i;
for (i = 0; i < n; ++i)
scanf("%d", &x[i]);
}

void output(int x[], int n)
{
int i;
for (i = 0; i < n; ++i)
printf("%d ", x[i]);
printf("\n");
}

int find_max(int x[], int n)
{
int m=x[0],i;
for(i=1;i<n;++i)
if(x[i]>m) m=x[i];
return m;
}

#include <stdio.h>
void dec2n(int x, int n);
int main()
{
int x;
printf("输入一个十进制整数: ");
scanf("%d", &x);
dec2n(x, 2);
dec2n(x, 8);
dec2n(x, 16);
return 0;
}

void dec2n(int x, int n)
{
int a[99],i=0,result;
while(x>0)
{
a[i]=x%n;
x=x/n;
i++;
}
for(i-=1;i>=0;i--)
{
switch(a[i])
{
case 10: printf("A");break;
case 11: printf("B");break;
case 12: printf("C");break;
case 13: printf("D");break;
case 14: printf("E");break;
case 15: printf("F");break;
default: printf("%d",a[i]);
}
}
printf("\n");
}

#include<stdio.h>
int main()
{
int n;
while(printf("Enter n:"),scanf("%d",&n)!=EOF)
{
int x[n][n],i,j,r,c;//r为行，c为列

for(r=0;r<n;r++)
for(c=0;c<n;c++)
x[r][c]=1;

for(r=0,c=0;r<n;r++,c++)
for(i=0;i<r;i++)
for(j=0;j<c;j++)
x[i][j]+=1;

for(r=n-1;r>=0;r--)
{
for(c=n-1;c>=0;c--)
printf("%d  ",x[r][c]);
printf("\n");
}
}
return 0;
}

学会数组的应用和二维数组

posted @ 2021-12-04 16:20  lzxy71  阅读(9)  评论(1编辑  收藏  举报